Problem: Rakesh and Tessa were asked to find an explicit formula for the sequence $100\,,\,50\,,\,25\,,\,12.5,...$, where the first term should be $f(1)$. Rakesh said the formula is $f(n)=100\cdot\left(\dfrac{1}{2}\right)^{{n-1}}$, and Tessa said the formula is $f(n)=200\cdot\left(\dfrac{1}{2}\right)^{{n}}$. Which one of them is right? Choose 1 answer: Choose 1 answer: (Choice A) A Only Rakesh (Choice B) B Only Tessa (Choice C) C Both Rakesh and Tessa (Choice D) D Neither Rakesh nor Tessa
Answer: In a geometric sequence, the ratio between successive terms is constant. This means that we can move from any term to the next one by multiplying by a constant value. Let's calculate this ratio over the first few terms: $\dfrac{12.5}{25}=\dfrac{25}{50}=\dfrac{50}{100}={\dfrac{1}{2}}$ We see that the constant ratio between successive terms is ${\dfrac{1}{2}}$. In other words, we can find any term by starting with the first term and multiplying by ${\dfrac{1}{2}}$ repeatedly until we get to the desired term. Let's look at the first few terms expressed as products: $n$ $1$ $2$ $3$ $4$ $f(n)$ ${100}\cdot\!\left({\dfrac{1}{2}}\right)^{0}$ ${100}\cdot\!\left({\dfrac{1}{2}}\right)^{1}$ ${100}\cdot\!\left({\dfrac{1}{2}}\right)^{2}$ ${100}\cdot\!\left({\dfrac{1}{2}}\right)^{3}$ We can see that every term is the product of the first term, ${100}$, and a power of the constant ratio, ${\dfrac{1}{2}}$. Note that this power is always one less than the term number $n$. This is because the first term is the product of itself and plainly $1$, which is like taking the constant ratio to the zeroth power. Thus, we arrive at the following explicit formula (Note that ${100}$ is the first term and ${\dfrac{1}{2}}$ is the constant ratio): $f(n)={100}\cdot\left({\dfrac{1}{2}}\right)^{{\,n-1}}$ So Rakesh is definitely right. What about Tessa? We can see that in Tessa's formula, the constant ratio is taken to the $n^{\text{th}}$ power. Let's expand the power in Rakesh's formula to arrive at a similar expression form: $\begin{aligned} f(n)= &{100}\cdot\left({\dfrac{1}{2}}\right)^{{\,n-1}}\\\\ = & 100\cdot\left(\dfrac{1}{2}\right)^{{n}}\cdot \left(\dfrac{1}{2}\right)^{{-1}}\\\\ = & 100\cdot \left(\dfrac{1}{2}\right)^{{n}}\cdot2\\\\ = & 200\cdot\left(\dfrac{1}{2}\right)^{{n}}\end{aligned}$ We obtained Tessa's formula, which means it's also a correct explicit formula for $f(n)$. Both Rakesh and Tessa got a correct explicit formula.